QJEEM24Ph2fdd: JEE Main 2024 Physics - Class 11 JEEM24Ph2fdd Solution: In a system two particles of masses and are placed at certain…

To solve this problem, we can make use of the concept of center of mass. The center of mass (CM) of a system remains unchanged if the internal forces act within the system without any external force. When one mass moves toward the CM, to keep the CM at the same position, the other mass must move in a way that the product of each mass with its displacement relative to the CM remains constant.

The formula to ensure that the center of mass remains unchanged can be derived from the principle of conservation of momentum or simply by understanding that the weighted average position (considering masses as weights) does not change.

Let's denote: - $x_1$ as the distance moved by $m_1$ towards the CM, - $x_2$ as the distance $m_2$ needs to move towards the CM, - The total mass of the system as $M = m_1 + m_2$ .

Since $m_1$ moves towards the CM by 2 cm, we apply the principle that the weighted sum of displacements (taking mass into account) remains 0 to maintain the center of mass at its original position:

$m_1 \cdot x_1 + m_2 \cdot x_2 = 0$

Given that $m_1 = 3 \, \text{kg}$ , $m_2 = 2 \, \text{kg}$ , and $x_1 = 2 \, \text{cm}$ , we substitute these values into the equation:

$3 \cdot 2 + 2 \cdot x_2 = 0$

Solving for $x_2$ gives:

$6 + 2x_2 = 0$

$2x_2 = -6$

$x_2 = -3 \, \text{cm}$

This means the mass $m_2$ should move $3 \, \mathrm{cm}$ towards the center of mass to keep the center of mass of the system at the original position. The negative sign indicates the direction is towards the center of mass, similar to $m_1$ 's movement direction in relation to keeping the CM stationary.